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I have the next positive definite matrix $Q\in\mathbb{R}^{3\times 3}$. For a full rank $U$ (e.g. defining a linear coordinate transformation) I can decompose $Q$ as $$ Q = U^{-1}TU, $$ where obviously $Q$ and $T$ are similar, but $T$ in general is not symmetric and then it is not positive definite.

Consider the rotation matrix $$ R(\theta) = \begin{bmatrix}sin(\theta) & cos(\theta) & 0 \\ -cos(\theta) & sin(\theta) & 0 \\ 0 & 0 & 1\end{bmatrix}, $$ and decompose $T$ as $$ T = R^T\tilde TR, $$ then we have the next relation $$ Q = U^{-1}R^T\tilde TRU. $$ And now decompose $U = R^T\tilde UR$, and therefore $U^{-1}=R^T\tilde U^{-1}R$. Substituting into the before equation we have $$ Q = R^T\tilde U^{-1} \tilde T \tilde U R \\ RQR^T = \tilde U^{-1} \tilde T \tilde U. $$ We know that $Q$ is similar to $\tilde U^{-1} \tilde T \tilde U$. So If $RQR^T$ is symmetric then $\tilde U^{-1} \tilde T \tilde U$ is positive definite.

But I cannot see why in general, for whatever rotation matrix (or even a more general set, orthogonal transformations?) $RQR^T$ is a symmetric matrix. My numerical simulations suggest that this is indeed the case.

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Use the property

$$(AB)^t=B^tA^t$$

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  • $\begingroup$ ohh, it was so easy! thanks! $\endgroup$ – user51196 Jun 9 '15 at 13:42
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Using tensor notation$$(RQR^T)_{ij}=R_{ik}Q_{kl}R^T_{lj}=R_{ik}Q_{kl}R_{jl}=R_{ik}Q_{lk}R_{jl}=R_{jl}Q_{lk}R_{ik}=R_{jl}Q_{lk}R^T_{ki}=(RQR^T)_{ji}$$

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